Electric potential is a scalar quantity that helps us understand the behaviour of charges in terms of energy. The charge possessed by an object and the relative position of an object related to other electrically charged objects is the two elements that give an object its electric potential energy.

We all know that positive charges flow from a higher potential to a lower potential, but how is the potential defined? Also, is the potential defined for a charge, or it is defined for a point? Will the potential of a point change if the magnitude of the charge we are bringing to that particular point changes? How is electric potential related to work done to move charges from one point to another? Let us read on to find out.

Electric Potential

Electric potential at a point is defined as work done per unit charge in order to bring a unit positive test charge from infinity to that point slowly. Here we assume the potential at infinity to be zero.
\(V_\infty = 0\)
The expression for an electric potential in terms of electric field can be derived as follows.
\({V_A} – {V_\infty } = \frac{{{W_{{\text{ext}}}}}}{{{q_0}}}\left( {\infty \to A,\,{\text{slowly}}} \right)\)
\({W_{{\text{ext}}}}\left( {\infty \to A} \right) = – \int_\infty ^A {\overrightarrow F } \cdot \overrightarrow {dr} \)
\({W_{{\text{ext}}}}\left( {\infty \to A} \right) = – \int_\infty ^A {{q_0}\overrightarrow E } \cdot \overrightarrow {dr} \)
\({V_A} = {V_A} – {V_\infty } = – \int_\infty ^A {\overrightarrow E } \cdot \overrightarrow {dr} \)
Electric potential is a scalar quantity, but it can be negative depending on the nature of the charge. Its SI unit is Volts \((\rm{V}).\)
Electric potential difference between two points ‘\(A\)’ and ‘\(B\)’ is defined as the work done per unit charge in moving a unit positive test charge from point ‘\(A\)’ to ‘\(B\)’ slowly.
\({V_B} – {V_A} = \frac{{{W_{{\text{ext}}}}}}{{{q_0}}}\left( {A \to B,\,{\text{slowly}}} \right)\)
\({W_{{\text{ext}}}}\left( {A \to B} \right) = – \int_A^B {\overrightarrow F } \cdot \overrightarrow {dr} \)
\({W_{{\text{ext}}}}\left( {A \to B} \right) = – \int_A^B {{q_0}\overrightarrow E } \cdot \overrightarrow {dr} \)
\({V_B} – {V_A} = – \int_A^B {\overrightarrow E } \cdot \overrightarrow {dr} \)
Here the point \(A\) is the reference point or initial point.
Conventionally we consider the point at infinity to be the reference point.
From the above definition, we can also infer that the magnitude of the electric field in a given direction is equal to the rate of change of potential with respect to distance,
\(\left| {\overrightarrow E } \right| = \frac{{dV}}{{dr}}\)

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Calculation of Electric Potential

Due to a Point Charge

A point charge is present in the space, and we need to calculate the potential at a point at a distance ‘\(r\)’, near the charge.
We know that the electric potential of a point is given by,
\({V_p} = {V_p} – {V_\infty } = – \int_\infty ^P {\overrightarrow E } \cdot \overrightarrow {dr} \)
Conventionally,
\(V_\infty = 0\)
\({V_p} = – \int_\infty ^P {\overrightarrow E } \cdot \overrightarrow {dr} \)
Putting the value of electric field due to a point charge we get,
\({V_p} = – \int_\infty ^r {\frac{q}{{4\pi {\varepsilon _0}{r^2}}}} \cdot \overrightarrow {dr} \)
\({V_p} = \frac{q}{{4\pi {\varepsilon _0}r}}\)
Due to the distribution of charges.
Potential is a scalar quantity, and therefore, the potential due a distribution of charges is given by the scalar sum of the potential due to individual charges.
\(V_P = V_1 + V_2 + V_3\)

Due to Continuous Charge Bodies

For continuous bodies, we get the potential by integrating the potential due to differential elements.

\(V_P = \int {dV} \)

Electric Potential due to a Charged Thread

\(V = \frac{\lambda }{{4\pi {\varepsilon _0}}}{\text{ln}}\left( {\frac{{{\text{sec}}\left( {{{{\theta }}_1}} \right) + {\text{tan}}\left( {{{{\theta }}_1}} \right)}}{{{\text{sec}}\left( {{{{\theta }}_2}} \right) – {\text{tan}}\left( {{{{\theta }}_2}} \right)}}} \right)\)
Where,
\(θ_1\) and \(θ_2\) are the angle subtended by the ends of the thread at that point as shown in the figure.
\(λ\) is the charge per unit length of the thread.
For an infinite charged thread, the potential at infinity is not zero, therefore in that case, the potential of a point is not defined, but the potential difference between two points can be calculated, which is given by,
\({V_B} – {V_A} = \frac{\lambda }{{2\pi {\varepsilon _0}}}{\text{ln}}\left( {\frac{{{r_A}}}{{{r_B}}}} \right)\)
Where,
\(r_A\) and \(r_B\) are the distance of the two points from the thread.

Potential due to a Charged Circular Arc

\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}R}}\)
Where,
\(R\) is the radius of the arc.
\(Q\) is the charge of the arc.

Potential due to a Charged Ring at a Point on its Axis

\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}\sqrt {{R^2} + {x^2}} }}\)
Where,
\(Q\) is the charge on the ring.
\(x\) is the distance of the point on its axis from the centre of the ring.
\(R\) is the radius of the ring

Potential due to Uniformly Charged Disc at a Point on its Axis

\(V = \frac{\sigma }{{2{\varepsilon _0}}}\left( {\sqrt {{R^2} + {x^2}} – x} \right)\)
Where,
\(σ\) is the area charge density of the disc.
\(x\) is the distance of the point on its axis from the centre of the disc.
\(R\) is the radius of the disc
Electric potential due to a uniformly charged shell
1. Outside the shell
\(r > R\)
\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}r}}\)
2. On the boundary of the shell
\(r = R\)
\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}R}}\)
3. Inside the shell
\(r < R\)
\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}R}}\)
The net electric field inside the shell is zero; therefore, the potential remains constant inside the shell.

Electric Potential due to Uniformly Charged Sphere

1. Outside the shell
\(r > R\)
\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}r}}\)
2. On the boundary of the shell
\(r = R\)
\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}R}}\)
3. Inside the shell
\(r < R\)
\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}{R^3}}}\left[ {\frac{3}{2}{R^2} – \frac{{{r^2}}}{2}} \right]\)

Electrostatic Potential Energy

Electrostatic potential energy can be defined as the work done by an external agent in changing the configuration of the system slowly.
Example: Three charges \(q_1,\;q_2\) and \(q_3\) are placed in space, and we need to calculate the electric potential energy of the system.
At first, we bring the first charge from infinity to origin. Since there is no electric field in space, the work done required to bring the first charge will be zero.
For the second charge, since the electric field is present due to the first charge, the work done to bring it from infinity to a point at a distance \(r_{12}\) is given by,
\(W_{ext} = qV\)
\( \Rightarrow {W_{ext}} = {q_2}\frac{{{q_1}}}{{4\pi {\varepsilon _0}{r_{12}}}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r_{12}}}}\)
For the third charge, we have an electric field due to two charges \(q_1\) and \(q_2\) present in space, thus work done in bringing the charge from the infinity to that point will be,
\({W_{ext}} = qV\)
\( \Rightarrow {W_{ext}} = \frac{{{q_1}{q_3}}}{{4\pi {\varepsilon _0}{r_{13}}}} + \frac{{{q_2}{q_3}}}{{4\pi {\varepsilon _0}{r_{23}}}}\)
Therefore, the net work done will be,
\(U = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r_{12}}}} + \frac{{{q_1}{q_3}}}{{4\pi {\varepsilon _0}{r_{13}}}} + \frac{{{q_2}{q_3}}}{{4\pi {\varepsilon _0}{r_{23}}}}\)
Thus, we can infer that the electric potential energy for distribution of charge will be,
\(U = \mathop {\sum \frac{{{q_i}{q_j}}}{{4\pi {\varepsilon _0}{r_{ij}}}}}\limits_{i \ne j} \)

Earthing

We consider Earth to be at zero potential, and if a conductor is connected to the Earth, then the potential of that conductor is also zero. Earth is always considered to be neutral, and therefore even if a large quantity of charge flows to the Earth the net charge will still remain unchanged, that is zero.

Solved Problems

Q.1. If the potential is defined in space to be,
\(V\left( {r,\,\theta } \right) = {r^2}{\text{sin}}\left( \theta \right)\)
Then find the magnitude of electric field at a point \(\left( {2\,{\text{m,}}\frac{\pi }{6}} \right).\)
Ans: We know that the magnitude of the electric field in a given direction is equal to the rate of change of potential with respect to distance,
\(\left| {\vec E} \right| = \frac{{dV}}{{dr}}\)
Here, we are using polar co-ordinates thus, we will have,
Electric field varying with angle and distance,
\({E_r} = – \frac{{\partial V}}{{\partial r}}\)
\({E_r} = \frac{{ – \partial V}}{{\partial r}} = – 2r{\text{sin}}\left( \theta \right)\)
And
\({E_\theta } = – \frac{{\partial V}}{{r\partial \theta }}\)
\({E_\theta } = \frac{{ – \partial V}}{{r\partial \theta }} = – r{\text{cos}}\left( \theta \right)\)
To find the magnitude of the net field,
\({E_{{\text{net}}}} = \sqrt {{E_r}^2 + {E_\theta }^2} \)
\( \Rightarrow {E_{{\text{net}}}} = r\sqrt {4{\text{si}}{{\text{n}}^2}\left( \theta \right) + {\text{co}}{{\text{s}}^2}\left( \theta \right)} \)
\( \Rightarrow {E_{net}} = 2\sqrt {3{\text{si}}{{\text{n}}^2}\left( {\frac{\pi }{6}} \right) + 1} = \sqrt 7 \;{\text{N}}\;{{\text{C}}^{ – 1}}.\)

Q.2. In the given figure, if the charge \(q_0\) is moved from \(A\) to \(B\), then find the change in electric potential energy of the system.

Ans: We know that,
Potential due to a charged ring at a point on its axis.
\({V_p} = \frac{Q}{{4\pi {\varepsilon _0}\sqrt {{R^2} + {x^2}} }}\)
Where,
\(Q\) is the charge on the ring.
\(x\) is the distance of the point on its axis from the centre of the ring.
\(R\) is the radius of the ring
The change in electric potential energy will be equal to the work done in moving the charge from \(A\) to \(B\),
\({U_f} – {U_i} = {W_{{\text{ext}}}} = {q_0}\left( {{V_A} – {V_B}} \right)\)
Potential at a point will be equal to the scalar sum of potential due to the two rings,
Thus, the potential at \(A\) is equal to,
\({V_A} = \frac{q}{{4\pi {\varepsilon _0}R}} – \frac{q}{{4\pi {\varepsilon _0}2R}} = \frac{q}{{4\pi {\varepsilon _0}2R}}\)
Potential at \(B\) will be,
\({V_B} = – \frac{q}{{4\pi {\varepsilon _0}R}} + \frac{q}{{4\pi {\varepsilon _0}2R}} = – \frac{q}{{4\pi {\varepsilon _0}2R}}\)
Change in electric potential energy will be,
\({U_f} – {U_i} = {W_{ext}} = {q_0}\left( {{V_A} – {V_B}} \right)\)
\( \Rightarrow {U_f} – {U_i} = {q_0}\left[ {\left( { – \frac{q}{{4\pi {\varepsilon _0}2R}}} \right) – \left( {\frac{q}{{4\pi {\varepsilon _0}2R}}} \right)} \right] = \frac{{ – q{q_0}}}{{4\pi {\varepsilon _0}R}}.\)

Summary

Electric potential of a point is defined as the work done per unit charge in bringing that charge from infinity to that point.
Electric potential is defined for a point, and it is independent of the magnitude of the charge kept at that point.
Electric potential is a scalar quantity, but it can take negative values depending on the direction of the electric field.
Work done in bringing a charge from one point to the other is given by,
\({W_{{\text{ext}}}} = {q_0}\left( {{V_A} – {V_B}} \right)\)
Where,
\(q_0\) is the value of the charge.
\(V_A\) and \(V_B\) are the potentials of the two points.

FAQs

Q.1. What are the electric potential difference and Electric potential?
Ans: Electric potential difference is defined between two points, that is, the work done per unit charge to bring the charge from one point to the other, whereas in the case of the electric potential, the initial point from which the charge is brought is located at infinity.
Electric potential difference is given by,
\({V_B} – {V_A} = \frac{{{W_{{\text{ext}}}}}}{{{q_0}}}\left( {A \to B,\,{\text{slowly}}} \right)\)
Electric potential is given by,
\({V_p} = {V_p} – {V_\infty } = \int_\infty ^P {\overrightarrow E } \cdot \overrightarrow {dr} \)
Conventionally,
\({V_\infty } = 0\)
\({V_p} = \int_\infty ^P {\overrightarrow E } .\overrightarrow {dr} .\)

Q.2. Is Electric potential a scalar quantity?
Ans: Yes, an electric potential is the dot product of the electric field and the distance. Therefore, it is a scalar quantity, but it can also be negative depending on the direction of the electric field.

Q.3. What is the potential of the Earth?
Ans: Earth is considered to always be at zero potential.

Q.4. What is electric potential energy?
Ans: Electric potential energy is defined for a system. It is the work done required to achieve the given configuration of the system, and it is a scalar quantity and depends on the magnitude and the distance between the charges on the system.

Q.5. What is the electric potential energy of a point charge kept at the origin?
Ans: We know that the electric potential energy is the work done to achieve the given configuration and since initially there is no electric field in space. Therefore, there is no work done in bringing the charge from infinity to the origin. Thus, the electric potential energy will be zero.

NCERT Solutions For Class 10 Science Chapter 12

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